Question

In a random sample of 360 women, 65% favored stricter gun control laws. In a random sample of 220 men, 60% favored stricter gun control laws. Test the claim that the proportion of women favoring stricter gun control is higher than the proportion of men favoring stricter gun control. Use a significance level of 0.05.

The test statistic is:    (rounded to 2 decimal places)

The p-value is:   (rounded to 4 decimal places)

The Conclusion

Answer 1

p1cap = X1/N1 = 234/360 = 0.65
p1cap = X2/N2 = 132/220 = 0.6
pcap = (X1 + X2)/(N1 + N2) = (234+132)/(360+220) = 0.631

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p1 = p2
Alternate Hypothesis, Ha: p1 > p2

Test statistic
z = (p1cap - p2cap)/sqrt(pcap * (1-pcap) * (1/N1 + 1/N2))
z = (0.65-0.6)/sqrt(0.631*(1-0.631)*(1/360 + 1/220))
z = 1.21

P-value Approach
P-value = 0.1131
As P-value >= 0.05, fail to reject null hypothesis.

There is not sufficient evidence to concldue that the proportion is higher in women